(Refer to steps 2.1-2.6) An experimenter sets up a microscope with a 5.33 mm obj
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Question
(Refer to steps 2.1-2.6) An experimenter sets up a microscope with a 5.33 mm objective lens. The object distance between the object and the center of the objective lens, d0 = 5.51 mm. What is The image distance (units required) between the objective lens and the image created by the objective, di = The tube length (units required) L = The lateral magnification of the objective M0 = If an eyepiece of focal length 125 mm is used, what is the magnifying power of this microscope? Calculated microscope magnifying power:Explanation / Answer
for objective lens ::
focal length = f = 5.33 mm
object distance = do = 5.51 mm
Using the lens equation ::
1/do + 1/di = 1/f
1/5.51 + 1/di = 1/5.33
di = 163.2 mm
focal length of eye piece = fe = 125 mm
tube length , L = di - f = 163.2 - 5.33 = 157.87 mm
magnification is given as ::
Mo = -di/do= -163.2 / 5.51 = -29.62
for eye piece ::
focal length of eye piece = fe = 125 mm
object distance = 125 mm
so using lens equation
1/fe = 1/do + 1/di
1/125 = 1/125 + 1/di
di = infinity
distance between the two lenses , d = di + fe = 163.2 + 125 = 288.2 mm
m = (L/fo )(1 + d/fe)
m = (157.87/5.33) (1 + 288.2/125)
m = 97.91
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